Introduction

In this blog post, we will explore how to solve the LeetCode problem "2483" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given the customer visit log of a shop represented by a 0-indexed string customers consisting only of characters 'N' and 'Y': if the ith character is 'Y', it means that customers come at the ith hour whereas 'N' indicates that no customers come at the ith hour. If the shop closes at the jth hour (0 <= j <= n), the penalty is calculated as follows: For every hour when the shop is open and no customers come, the penalty increases by 1. For every hour when the shop is closed and customers come, the penalty increases by 1. Return the earliest hour at which the shop must be closed to incur a minimum penalty. Note that if a shop closes at the jth hour, it means the shop is closed at the hour j. Example 1: Input: customers = "YYNY" Output: 2 Explanation: - Closing the shop at the 0th hour incurs in 1+1+0+1 = 3 penalty. - Closing the shop at the 1st hour incurs in 0+1+0+1 = 2 penalty. - Closing the shop at the 2nd hour incurs in 0+0+0+1 = 1 penalty. - Closing the shop at the 3rd hour incurs in 0+0+1+1 = 2 penalty. - Closing the shop at the 4th hour incurs in 0+0+1+0 = 1 penalty. Closing the shop at 2nd or 4th hour gives a minimum penalty. Since 2 is earlier, the optimal closing time is 2. Example 2: Input: customers = "NNNNN" Output: 0 Explanation: It is best to close the shop at the 0th hour as no customers arrive. Example 3: Input: customers = "YYYY" Output: 4 Explanation: It is best to close the shop at the 4th hour as customers arrive at each hour. Constraints: 1 <= customers.length <= 105 customers consists only of characters 'Y' and 'N'.

Explanation

Here's the breakdown of the solution:

• Calculate prefix sums for 'N' and 'Y': Precompute the cumulative count of 'N's from the beginning and 'Y's from the end of the string. This allows calculating penalties for different closing hours efficiently.
• Iterate through possible closing hours: Loop through each possible closing hour (0 to n) and calculate the penalty based on the prefix sums.

• Track minimum penalty and corresponding hour: Keep track of the minimum penalty encountered so far and the corresponding closing hour.

• Runtime Complexity: O(n), where n is the length of the customers string.

• Storage Complexity: O(n) due to the prefix sums.

Code

    def bestClosingTime(customers: str) -> int:
    n = len(customers)
    prefix_n = [0] * (n + 1)
    suffix_y = [0] * (n + 1)

    for i in range(n):
        prefix_n[i + 1] = prefix_n[i] + (1 if customers[i] == 'N' else 0)

    for i in range(n - 1, -1, -1):
        suffix_y[i] = suffix_y[i + 1] + (1 if customers[i] == 'Y' else 0)

    min_penalty = float('inf')
    best_hour = 0

    for i in range(n + 1):
        penalty = prefix_n[i] + suffix_y[i]
        if penalty < min_penalty:
            min_penalty = penalty
            best_hour = i

    return best_hour